Chapter 20: Basic principles of intersection signalization
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Chapter objectives: By the end of this chapter the student will be able to:
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Explain the meanings of the terms related to signalized intersections
Explain the relationship among discharge headway, saturation flow, lost times, and capacity
Explain the “critical lane” and “time budget” concepts
Model left-turn vehicles in signal timing
State the definitions of various delays taking place at signalized intersections
Graph the relation between delay, waiting time, and queue length
Explain three delay scenarios (uniform)
Explain the components of Webster’s delay model and use it to estimate delay
Explain the concept behind the modeling of random and overflow delay
Know inconsistencies existing between stochastic and overflow delay models
Chapter 20
1

Four critical aspects of signalized intersection operation discussed in this chapter
3.
4.
1.
2.
Discharge headways, saturation flow rates, and lost times
Allocation of time and the critical lane concept
The concept of left-turn equivalency
Delay as a measure of service quality
Chapter 20
2

Controller
Cycle length
Phase
Interval
Change interval
All-red interval
(clearance interval)
3
Chapter 20

Signal timing with a pedestrian signal: Example
% Interval
5
6
3
4
1
2
7
8
Veh.
G-26
Pine St.
Ped.
W-20
Y-3.5
R-25.5
FDW-6
DW-29
Veh.
R-31
Oak St.
Ped.
DW-31
G-19
Y-3
R-2
Cycle length = 55 seconds
Chapter 20
W-8
FDW-11
DW-5
36.4
10.9
6.4
AR 2.7
14.5
20.0
5.5
AR 3.6
4

20.1.2 Signal operation modes and left-turn treatments & 20.1.3 Left-turn treatments
Operation modes:
Pretimed (fixed) operation
Semi-actuated operation
Full-actuated operation
Master controller, computer control, adaptive traffic control systems for coordinated systems
Left-turn treatments:
Permitted left turns
Protected left turns
Protected/permitted
(compound) or permitted/protected left turns
Chapter 20
5

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LT flow rate
Opposing flow rate
Number of opposing lanes
Whether LTs flow from an exclusive LT lane or from a shared lane
Details of the signal timing
Chapter 20
6

Bangerter Highway &
3500 South
Chapter 20
7

Chapter 20
8

Four basic mechanisms for building an analytic model or description of a signalized intersection
Discharge headways at a signalized intersection
The “critical lane” and “time budget” concepts
The effects of LT vehicles
Delay and other MOEs (like queue size and the number of stops)
Chapter 20
9

20.2 Discharge headways, saturation flow, lost times, and capacity
Δ(i) Start-up lost time
Effective green h
1 2 3 4 5 6 7 s l
1
T


3600

 h

( i ) l
1
 nh
Saturation flow rate
Startup lost time
G i
Vehicles in queue
Total lost time e
Clearance lost time
Capacity g
Y i i


G i y i


Y i ar i t
L l
2

 y l
1

 l
2 ar

 t
L e c i
 s i g
C i
Extension of green y i ar i
Cycle length
Chapter 20
10

First approach:
Chapter 20
Second approach:
Eq. 20-6
11

Chapter 20
12

20.3 The “critical lane” and “time budget” concepts
Each phase has one and only one critical lane (the most intense traffic demand). If you have a 2-phase signal, then you have two critical lanes.
345
L
H

Nt
L
3600
C
Total loss in one hour
75
100
T
G
V c

3600

Nt
L

T h
G

3600
1 h
C

 3600

Nt
L
Total effective green in one hour
3600
C


450
Max. sum of critical traffic demand; this is the total demand that the intersection can handle.
N = No. of phases; t
L
= Lost time in seconds per phase; C = Cycle length, sec; h = saturation headway, sec/veh
Chapter 20
13

20.3.2 Finding an Appropriate Cycle Length
The benefit of longer cycle length tapers around 90 to 100 seconds. This is one reason why shorter cycle lengths are better.
N = # of phases. Larger N, more lost time, lower V c
.
Desirable cycle length, incorporating
PHF and the desired level of v/c
C min
C des


1

Nt
L
V c
Eq. 20-13
1

3600 / h
Nt
L
V c
Eq. 20-14
PHF ( v / c )( 3600 / h )
Doesn’t this look like the Webster model?
C
0
Y i


1 .
5 L
1

 i



1
Y i
5 flow _ ratio ( v / s ) i
Chapter 20
14

C
0

1

1 .
5 L

5 i



1
  i
C
0
= optimal cycle length for minimum delay, sec
L = Total lost time per cycle, sec
Sum (v/s) i
= Sum of v/s ratios for critical lanes
Delay is not so sensitive for a certain range of cycle length

This is the reason why we can round up the cycle length to, say, a multiple of 5 seconds.
15
Chapter 20

20.3.2 Finding an Appropriate Cycle Length
Desirable cycle length, C des
Cycle length
100% increase
Fig. 20.4
V c
8% increase
Chapter 20
Marginal gain in
V c decreases as the cycle length increases.
(Review the sample problem on page 473)
16

V c

T
G h

1 h

 3600

Nt
L
3600
C


C des

1

Nt
V
L c
PHF ( v / c )( 3600 / h )
17
Chapter 20

20.4 The Concept of Left-Turn (and Right-
Turn) Equivalency
In the same amount of time, the left lane discharges 5 through vehicles and 2 left-turning vehicles, while the right lane discharges 11 through vehicles.
5

2 E
LT

11 and :
E
LT

11

5
2

3 .
0
Chapter 20
18

5
1000 1500
The LT equivalent increases as the opposing flow increases.
For any given opposing flow, however, the equivalent decreases as the number of opposing lanes is increased.
Chapter 20
19

Given conditions:
 2-lane approach
 Permitted LT
 10% LT, TVE ( E
LT
) =5
 h = 2 sec for through
Solution 1: Each LT consumes 5 times more effective green time.
h prev

( 0 .
1 )( 5

2 .
00 )

( 0 .
9 )( 2 .
00 )

2 .
80 sec/ h s

3600

3600 h prev
2 .
80

1286 vphgpl
Solution 2: Calibrate a factor that would multiply the saturation flow rate for through vehicles to produce the actual saturation flow rate.
s

1800 ( 0 .
714 )

1286 vphgpl f
LT s
 3600
2

1800 vphgpl
 h ideal h prev


P
LT
E
LT h ideal h ideal  
( 1

P
LT
)( 1 .
0 ) h ideal
 or s

1800 ( 2 .
0 / 2 .
8 )

1800 ( 0 .
714 )

1286

1

1
P
LT
( E
LT

1 )

1

1
0 .
10 ( 5

1 )

0 .
714
Chapter 20
20

Common MOEs:
• Delay
• Queuing
• No. of stops (or percent stops)
Stopped time delay: The time a vehicle is stopped while waiting to pass through the intersection
Approach delay: Includes stopped time, time lost for acceleration and deceleration from/to a stop
Travel time delay: the difference between the driver’s desired total time to traverse the intersection and the actual time required to traverse it.
Time-in-queue delay: the total time from a vehicle joining an intersection queue to its discharge across the stopline or curb-line.
Control delay: time-in-queue delay + acceleration/deceleration delay)
Chapter 20
21

Uniform arrival rate assumed, v
Note that W(i) is approach delay in this model.
Here we assume queued vehicles are completely released during the green.
At saturation flow rate, s
Figure 20.10
Chapter 20
The area of the triangle is the aggregate delay.
22

This is acceptable.
This is great.
UD = uniform delay
OD = overflow delay due to prolonged demand > supply
(Overall v/c > 1.0)
OD = overflow delay due to randomness (“random delay”).
Overall v/c < 1.0
If this is the case, we have to do something about this signal.
Chapter 20
A(t) = arrival function
D(t) = discharge function
23

Isolated intersections
Signalized arterials
HCM uses the Arrival Type factor to adjust the delay computed as an isolated intersection to reflect the platoon effect on delay.
Chapter 20
24

UD a
V
R

 v
C  1 c g
C

R
 t



 vR
 vt c
 st c t
V c
UD a

1
2
( base : R )( height : V )

1
2
C
2
 1 g
C


2 
 s vs
 v


The area of the triangle is the aggregated delay, “Uniform Delay (UD)”.

 s vR
 v st c


C  1 s v
 v
C  1 g
C

 g
C



 s vs
 v


Total approach delay
To get average approach delay/vehicle, divide this by vC
UD

C
2

1

1


 g C
 



2
25
Chapter 20

D


C
0

2

1

1


 g C
 



2
.
65

2 v

1
 

2
  
 
1 3
 
2

 g C


UD = uniform delay
Adjustment term for overestimation
(between 5% and 15%)
Analytical model for random delay
OD = overflow delay due to randomness (in reality “random delay”). Overall v/c < 1.0
D = 0.90[UD + RD]
26
Chapter 20

Random delay derivation
Chapter 20
Chapter 20.
27

UD o


C
2
C

1

1

1



(
 g g C
 
C )




2
2

C
2

1


1

 g

/ g
C
/ C v
 
2
   because c = s (g/C), divide both sides by v and you get (g/C)(v/c) = (v/s).
And v/c = 1.0
.
The aggregate overflow delay is:
OD a

1
2
T
 vT
 cT


T  v
 c

2
2
Because the total vehicle discharged during T is cT ,
OD

T
2
  

1


T
2

X

1

See the right column of p.482 for the
28

1
2
OD

T
1

T
2
2
  

1

Average delay/vehicle =
(Area of trapezoid)/(No. vehicles within T
2
-T
1
).
Chapter 20
Derive it by yourself.
Hint: the denominator is c(T
2
-T
1
).
29

20.5.3 Inconsistencies in random and
D overflow delay


C
0

2

1

1


 g C
 



2
.
65

2 v

1
 

2
  
 
1 3
 
2

 g C


OD

T
2
  

1

The stochastic model’s overflow delay is asymptotic to v/c = 1.0 and the overflow model’s delay is 0 at v/c =1.0. The real overflow delay is somewhere between these two models.
30
Chapter 20

Comparison of various overflow delay model
20.5.4 Delay model in the HCM 2000
The 4 th edition dropped the HCM 2000 model (I don’t know why…). It looks like Akcelik’s model that you see in p. 484 (eq. 20-26).
These models try to address delays for 0.85< v/c <1.15 cases.
Chapter 20
31

We will walk through sample problems
(pages 484-485). This will review all delay models we studied in this chapter.
Start reading Synchro 9.0 User Manual and SimTraffic 9.0 User Manual. We will use these software programs starting Mon, October 20, 2014.
Chapter 20
32